A takes 2 hours more than B to cover a distance of 40 km. If A doubles his speed, he takes $1\frac{1}{2}$ hour more than B to cover 80 km. To cover a distance of 120 km, how much time(in hours) will B take travelling at his same speed?

$1 \frac{1}{2}$

Let speed of B = S km/h

Time taken by B to cover 40km  = \(\frac{40}{S}\)

Time taken by A = \(\frac{40}{S}\) + 2

Now , Distance = 80 km

Time taken by B to cover 80 km = \(\frac{80}{S}\)  --------(2)

Speed of A when he double his speed = \(\frac{40S}{20 + S}\)

Time taken by A to cover 80 km = \(\frac{40 + 2S}{S}\)  ------(3)

Subtracting (2) from (3) ,

 \(\frac{40 + 2S}{S}\)  -  \(\frac{80}{S}\)  =  \(\frac{3}{2}\) 

2 ( 40 + 2S - 80) = 3S

4S - 80 = 3S

S = 80

Speed of B = 80 km/h

Distance = 120 km

Time taken by B = \(\frac{120}{80}\) 

So , Time taken by B to cover 90 km is 1\(\frac{1}{2}\) hours.